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Question

Two identical charged spheres suspended from a common point by two massless strings of lengths $l$ , are initially at a distance $d (d << l)$ apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity $v$ . Which of the following correctly represent the variation of $v$ as a function of the distance $x$ between the spheres.

v \propto x^{\frac{1}{2}}

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v \propto x^{- \frac{1}{2}}

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v \propto x^{- 1}

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v \propto x^{- \frac{1}{2}}

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Solution

The correct option is B $v \propto x^{- \frac{1}{2}}$

From figure,

$T cos θ = m g \dots \dots (1)$
$T sin θ = \frac{k q^{2}}{x^{2}} \dots \dots (2)$

From eqns. ( $1$ ) and ( $2$ ),

$tan θ = \frac{k q^{2}}{x^{2} m g}$

Since $d << l$ so will be $x << l$ . so $θ$ is small,
$∴ tan θ \approx sin θ = \frac{x}{2 l}$
$∴ \frac{x}{2 l} = \frac{k q^{2}}{x^{2} m g} \Rightarrow q^{2} = x^{3} \frac{m g}{2 l k}$

$\Rightarrow q \propto x^{\frac{3}{2}}$

Differentiating with respect to time on both sides we get,

$\frac{d q}{d t} \propto \frac{3}{2} \sqrt{x} \frac{d x}{d t}$

$\Rightarrow \frac{d q}{d t} \propto \frac{3}{2} \sqrt{x} v$

From the data given in the question,

$\frac{d q}{d t} = constant$

$∴ v \propto \frac{1}{\sqrt{x}}$

Hence, option (a) is the correct answer.

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Spread of the Mutiny

Standard X History

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